radial bg, to cut the side bc or b c produced. Join B 5. Produce the side c d, and draw lines from 6 and 7, parallel to cg, cutting cd or cd produced. Join g 6 and g 7. But if, as in this case, the last division falls beyond the side c d, draw a line from the point on the side cd, parallel to the next radial, dg, to cut the side de in o, and join o'g. Then these lines go, g 1, g 2, g 3, g 4, g 5, and g 6, will divide the figure abcdef into the required parts or areas. FIG. 73.

All parallelograms, whether right or oblique angled, having equal bases and equal altitudes, have equal areas.

ABCD is a right-angled parallelogram or rectangle. ABEF is an oblique-angled parallelogram. The base A B and perpendicular height B D is common, or the same for both; therefore A B C D is equal to A BEF. The superficial areas of all parallelograms are obtained by multiplying their base by their height. All parallelograms having equal altitudes have to each other the same ratios (or proportion) that their bases have. Again, all parallelograms having equal bases have to each other the same ratios that their altitudes have.

FIG. 74.

All triangles having equal bases and equal altitudes, have equal areas.

The right-angled triangle ABC has the same base and altitude as triangle A B D, and also triangle A BE; therefore, their areas are equal. The superficial areas of all triangles are obtained by multiplying their base by half their altitude (or perpendicular height), or their altitude by half the base.

All triangles having equal altitudes, are to each other in the same ratio as their bases; or, having equal bases, are in the same ratio as their altitudes.

FIG. 75.

To draw a triangle equal in area to a rectangle; or to draw a rectangle equal to a given triangle.

Let A B C be a given rectangle. Make any triangle on A B having twice B C for its perpendicular height, as A B D. Then triangle A B D will be equal in area to the given rectangle A B C. If the triangle A B D be given, take half its perpendicular height for the height B C of the rectangle, and make the base of the rectangle equal to that of the triangle. Then the rectangle will be equal in area to the given triangle.

NOTE. Therefore, any triangle drawn on the base of a rectangle, and having twice the perpendicular height, or having the same height and twice the base, is equal to it in area, and vice versa. Any rectangle having the same base as a triangle and half the height, or half the base and the whole height, is equal to it in area.

FIG. 76.

To draw a square equal in area to any given rectangle.

Let A B C be a given rectangle. Produce A B to C', making B C equal B C. On D, as a centre, draw a semicircle A F E C. Produce the side B C to cut the semicircle E. Then B E will be a mean proportional, as in Figure 46, between the two sides of the given rectangle (A B and B C), on which construct a square B E F G.

FIG. 77.

To draw any rectangle equal in area to a given square.

Let A B C be a given square. Produce the side A B indefinitely, and determine the length of one side of the required rectangle, as B F. On E, as a centre, with a suitable radius, draw a semicircle to pass through F and C, cutting the base-line in D. Draw the rectangle B F G, making F G equal to B D. Then B F G will be the required rectangle, equal in area to the given square. In the same way, if B D be given to find B F, draw the semicircle through DC to cut A B produced in F.

The second side of the rectangle may be obtained by Fig. 47, finding a third proportional between the given side of the rectangle and square.

FIG. 78.

To draw any rectangle equal in area to a given one, but having a different base or height.

Let A B C be a given rectangle, and D E the base of the required one. Find E F, a fourth proportional (Figure 48) to A B, B C, and D E. In the following manner, draw a line from D, making any angle (about 45°) with D E. Mark off a distance on it equal to A B. On D E mark off a distance equal to B C, from D towards E. Join the ends of the two lines, and draw a line parallel to it from the intermediate point in D E, to cut the other line at a distance from D, equal to the height of the required rectangle, as E F. Then the rectangle D E F will be equal to the given rectangle A B C. If the side E F is given to find D E, proceed as before; reversing the process to find D E, the fourth proportional greater. FIG. 79.

To draw a triangle equal to a given one, but having a different base or altitude. Let A B C be the given triangle, and D the altitude of the required one. Produce one side, as A C, to cut line D. Join D B, and draw a line parallel to it from C to cut the base A B in E. Draw the diagonal D E, which will be the side of the triangle required. Then triangle A E D will be equal to triangle A B C. If the base, A E, is given to find the height D, join C E and draw a line parallel to it from B, to cut the produced side A C in D. Draw line ED as before. This is only the reverse of the foregoing.

NOTE. This figure may also be worked by a fourth proportional to the height and base of the given triangle, and the height or base of the required one, as in the previous figure.

FIG. 80.

To draw a triangle equal in area to a given one, and similar to another.

Let a b c be the given triangle, representing the given area; and A the form of the required triangle. Change triangle a b c into one (a e d) of the same height as A by the previous figure. Find a mean proportional, fg, between the base of the triangle a b d, and triangle A. On ƒg, as a base, construct a triangle similar to A. Then this triangle, C, will be equal to one, B, and similar to the other, A. A triangle may be made similar to a given one, and equal to any given figure or area by this method.

FIG. 81.

To draw an equilateral triangle equal in area to a given triangle not equilateral.

Let A B C be the given triangle. Draw an equilateral triangle upon one side of the given triangle, as A B D. Produce line D A. From C draw a line parallel to the side A B, and cutting the produced line, D A, in E. By Fig. 46 find a mean proportional, A G, between lines D A and A E; that is, describe a semicircle upon D E. At A, draw A G perpendicular to D E, and cutting the semicircle in G. Line A G will be the mean proportional and side of the required equilateral triangle, equal in area to the given triangle. Therefore, any parallelogram, polygon, circle, or trapezium, may be converted into an equilateral triangle having the same area, by first changing the given figure into any equal triangle, and then proceeding as described.

FIG. 82.

To draw a rectangle equal in area to a circle. Let A be the centre of the given circle. Draw a diameter through A, and also a radius A B, perpendicular to it, so as to find a quarter of the circumference B C. Draw a rectangle D G H I, making one side, D G, equal the diameter of the circle, and the other side, G H, equal B C, one quarter of its circumference. Or, make a rectangle D E F, having one side, E F, equal to the radius, A C, of the circle, and the other side, D E, equal to one half of its circumference. Then each of these rectangles, D G HI and D E F, are equal to the given circle. Therefore, a triangle may also be made equal to a circle, by making its height equal to the diameter, and its base equal to half the circumference.

area.

FIG. 83.

To draw a triangle of any given area.
Draw a rectangle, A B C D, to represent the given

This is done by making one side, A B, one unit in length, and the other side the same number of units or parts as the area. Thus, for an area of three and a half units, make A B 1 unit and B C 3 units. In some cases, it is found more convenient to take two or four units for its height and half or a quarter the area in units for its base. Erect any perpendicular, as E F, equal to twice the height of the rectangle; and join E B and E C. Then the triangle, BE C, will be equal to the given rectangle, and hence contain the given area. Therefore, a triangle may be made to contain any given area at once without drawing the rectangle, by making its base equal and its height double.

FIG. 84.

To draw a square of any given area. Make a rectangle A B C D, to represent the area, as in the previous figure. Find a mean proportional (ce) between the sides of the rectangle, B C and C D. Then a square made on c e for its side, will contain the area required.

FIG. 85.

To draw a rectangle equal to a given square, when one side of the rectangle is given.

Let A B C be the given square. Draw the given side of the rectangle, B E, on A B produced; then the

other side may be found by drawing a semicircle through E and C to cut A B, giving B D; or, by finding a third proportional to the side of the square and the given side of the rectangle.

FIG. 86.

The area of a circle is equal to a rectangle, when two of the sides are equal to half the circumference, and the other two to the radius of the circle.

If a circle, A B, be divided into a number of equal parts or sectors, as twenty-four, and twelve of them are opened out upon a straight line, A B, which will be equal to half the circumference (nearly)—the remaining twelve parts, being inverted, will, with the others, form a rectangle nearly equal to the circle (the error is about 800).

FIG. 87.

To draw a triangle equal in area to a given sector.

Let A B C D be a given sector. Find a rectangle, B C A, equal to the sector, by making C A equal the radius, and B C equal half the circumference of the sector. Then a triangle having its perpendicular height the same as, and its base double that of, the rectangle, will be equal to the given sector. FIG. 88.

See Fig. 79.

FIG. 89.

See Fig. 81.

FIG. 90.

See Fig. 80.

FIG. 91.

To draw a triangle equal to the sum of two or more given triangles.

Let A and B be the two given dissimilar triangles. Draw a triangle, a b c, similar to B. Produce the side ac to d. Change this triangle into one of the same height as A by Fig. 79-d e being the side of the new triangle (a de). Add the base of the triangle A to a c, which produces it to f. Join df. Then the triangle a d f will be equal to the sum of the two triangles. If there are more triangles to be added, change each into triangles of the same height as the first, and add their bases together.

FIG. 92.

To reduce any given figure to an equal triangle. Let A B C D E be the given figure, and the vertex of the triangle be required at D. Then join A C, and draw a line from B parallel to it, cutting the base produced to a. If a and C are joined, a C D E will be equal to the given figure. Join a D, and draw a line from C parallel to it, cutting the base produced to f. Join fD, then the triangle ƒD E will be equal to the given figure as required. And if the given figure has more sides, proceed in the same way, reducing the figure one side less each time, until it becomes a triangle. The vertex of the triangle may be in any of the corners or sides, and the base may be a continuation of any side.

FIG. 93.

To draw a triangle equal to a given lunule. Let A B be the given lunule. Join A B, the diameter of the outer arc, and find the diameter D E of the inner arc. From D as a centre, with the diameter of the outer arc, A B, as radius, cut the inner arc in

To draw a lunule equal to a given triangle. Let a b c be the given triangle, which must have one right angle. Draw a perpendicular to the longest side a b at b, cutting the other side produced in d. Draw a semicircle A B, with half a b as radius, then this will be the outer arc of the lunule. Take half a d as radius; and from a point C, draw an arc, cutting the other at A and B. This being the inner arc. The space D enclosed by the two circles forms the lunule which is equal to the given triangle.

To draw an ellipse equal to a given circle. Let A B be the given circle. Determine one axis or diameter of the ellipse, as D D. By Fig. 47 find a third proportional to half A B and half D D. Then C E, the third proportional, will be half the other diameter of the ellipse required; and the curve being drawn by the rule of Fig. 54, this ellipse will be equal in area to the given circle. It is obvious that either the half or the whole diameters may be taken to find the third proportional.

FIG. 97.

To draw any regular polygon equal to a given triangle. Let A B C be any given triangle. Divide one side of the triangle (as A B) into the same number of equal parts as there are to be sides to the polygon-in this case, six. Draw a line A E, forming an angle with the other side A C, of one-sixth of 360°, which is the angle at the centre of the required polygon. From the first division 1, draw a line parallel to A C, to cut A E in F. Find a mean proportional (A H) between A C and A F. Then A H will be the radius of the circumscribing circle, D E H, of the polygon. Produce AF to E, then A C and A E cut the circle into onesixth. Draw the chord of this part, which will form one side of the polygon. Complete the polygon by stepping this distance round the circle, and joining the points. FIG. 98.

To draw a parallelogram equal in area and perimeter to any given triangle.

Let A B C be the given triangle. Draw two indefinite parallel lines, a d and b c, at a distance from each other equal to the perpendicular height of the triangle from the base A C to the point B. Take half A C for the base of the parallelogram a d. From a and d as centres, with a radius E F equal to half the sum of the other two sides of the triangle, draw arcs cutting the line in b and c. Join a band a c. Then this parallelogram, a b cd, is equal to the triangle in area, because it has the same perpendicular height, and half the base of the triangle. The perimeter or

sum of the sides is equal to that of the triangle, because the two sides, a d and b c, are together equal to the base A C, and the two sides a b and d c are equal to the sum of the two other sides A B, B C. FIG. 99.

To draw a square equal to the sum of two or more given

squares.

Let A B and A C be the sides of two given squares D and E. Place these two sides at right angles to each other. Join the ends C and B. Then the square F, on B C, is equal to the squares on the other lines.

This rule is useful in adding any other similar figures together. FIG. 100.

To draw a square equal to the difference between two given

squares.

Let A B and B C be the sides of the two given squares. Draw A B, and at one end of it erect a perpendicular, A C. From B, with a radius B C, draw an arc to cut the perpendicular in C. Then A C is the side of a square equal in area to the difference between the two given ones. The difference between any two similar figures may be found by this means. FIG. 101.

To draw any figure having an area in any proportion, greater or less, than a given similar figure.

or

Let A B represent the base or side of a triangle parallelogram, or the diameter of a circle or polygon. 1st. If the proposed figure be a triangle, a b being its base, and it is required to find a triangle of onehalf its area. Produce A B indefinitely. Take half of A B, and set it off from B, as B. Draw a semicircle on the whole line A. At B erect a perpendicular, cutting the circle at D. Then B D will be the base of the new triangle, a c, required, shown by the line at c, drawn parallel to the side of the given triangle (so as to retain the same angles).

2nd. If the given figure is a parallelogram or rectangle, a b being one side, and a similar figure of one quarter the area be required. Draw A B, equal to a b, and add one quarter of A B to it; and on A as a diameter, draw a semicircle, cutting the perpendicular at B in C. Then B C is the side of the required rectangle a c. Draw the diagonal of the rectangle, and also a line from parallel to the second side of it. From the point where this side cuts the diagonal, draw another line parallel to the third side of the given figure. Then this inner figure will be similar, and contain one quarter the area of the

outer one.

3rd. If the given figure be any irregular polygon, and it is proposed to draw one similar, of one half the area, proceed as before. From A draw lines passing through all the angles, and from the point found, c, draw lines parallel to the sides of the figure, and cutting the radii, as shown.

4th. If the given figure is a circle or a regular polygon, of which a b is the diameter, let A B represent this diameter. It is proposed to draw a figure of 1 times the area of the given one. From B set off 1 times A B (as at 1). On A 1, as a diameter, describe a semicircle. Erect a perpendicular

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